Solving trigonometric equations

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as we know that (sin(x))^2 + (cos(x))^2=1
how about (sin(3x))^2 + (cos(3x))^2 and 5sin(3x)^2 + 6cos(3x)^2??

how can we solve these problems??
thanx
 
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What are you trying to solve? You could substitute numbers for the x's and compute the value.

I don't believe there's a nifty identity for (sin(3x))^2 + (cos(3x))^2 even though it does look somewhat similar to (sin(x))^2 + (cos(x))^2.
 
Actually, aslong as the arguments match it equals one. So (sin(3x))^2 + (cos(3x))^2 does equal one.
 
Indeed, if we apply the identity \sin^{2}\theta +\cos^{2}\theta = 1,to the above expression and simply use the substitution 3x = \theta... :wink:
 
For 5sin(3x)^2 + 6cos(3x)^2, write it as 5sin^3(3x)+ 5cos^2(3x)+ cos^2(3x)= 5(sin^3(3x)+ cos^2(3x))+ cos^2(3x)= 5+ cos^2(3x).
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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